Quiz

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The aim of any calculation in physics is to find an equation that relates the variables of the system. Such relations are incredibly useful because they capture the behavior of infinitely many instantiations of a problem in a single, comprehensible statement of equality. For example, we know how a bomb's blast radius grows over time, given the density of the atmosphere and the bomb's energy. We know how fast a projectile must be fired to escape the gravitational field of a planet. We can even estimate the amount of time needed for a stirred cup of coffee to settle. Because equalities are so useful, great time and intellectual investment is devoted to finding them through any number of mathematical techniques and calculating tricks.

What causes a cup of coffee to settle down?

All methods, however, are not equal, and there exists a trick, based on the simplest of observations, that can let us leapfrog from the bowels of confusion to a point where we have a valid equation that correctly describes a property of our system, without any calculation at all. The trick is dimensional analysis.

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Dimensional Analysis

The observation is that if an equation correctly describes the physics of a system, the least we can say is that each side of the equation has the same units as the other. In other words, we must be comparing apples to apples. We can take any well-known result, the equivalence of mass and energy, for instance, and find this to be the case. For a free particle, we have \(E=mc^2\), where \(c\) is the speed of light, \(m\) is its relativistic mass, and \(E\) is its total energy. On the right-hand side, we have units of mass multiplied twice by units of velocity, or \(\text{kg}\text{ m}^2 / \text{ s}^2\), which we recognize as Newton meters, \(\text{Nm}.\) On the left, we have energy, which has units of Newton meters by definition. Thus, we see that we have an equation relating energy to energy... big surprise! At the risk of further underwhelming, we point out that the quotient, \(\frac{E}{mc^2}\), is a dimensionless product, i.e. is a parameter that has no physical units, just a pure number. It is these dimensionless quantities that we are after because they are the only combinations of our variables that can possibly give rise to correct physical relationships.

Before we move on, it is useful to introduce twiddle notation, \(\sim\). If a quantity \(g\) is equal to some combination of factors \(f(r_1,\ldots,r_n)\), save for some numerical multiplier \(\gamma\) such that \(g=\gamma f(r_1,\ldots,r_n)\), then we say \(g\sim f(r_1,\ldots,r_n)\).

For example, if \(f = 24\pi l^2\), or \(f =\frac{ l^2}{100}\), we say in both cases that \(f\sim l^2\). For all \(f\) for which this relation holds, we say that \(f\) scales as \(l^2\). Up to some numerical constant that can be obtained by measuring \(f\) at a single value of \(l^2\), \(f\) is determined entirely through its dependence on the quantity \(l\).

Twiddle representation of famous results:

Some famous relations in math and physics, recast using twiddle notation, are as follows:

Another closely related notation is \(\propto,\) which is used to denote that the left-hand side is proportional to the right-hand side. In contrast to \(x\sim l\), \(x \propto l\) does not mean that the behavior of \(x\) is captured soley through \(l\), but that \(l\) is a factor. For instance, if \(f(x,y,z) = \pi xyz\), we can say \(f\propto x\), \(f\propto y\), and \(f\propto z\).

As we mentioned above, any true physical relation must have equal units on both sides. Thus, if we ignore dimensionless pre-factors (pure numbers), we have \(\displaystyle\text{LHS}\sim\text{RHS}\), or equivalently, \(\frac{\text{LHS}}{\text{RHS}}\sim 1\) for all physical relations. More generally, we can replace \(\frac{\text{LHS}}{\text{RHS}}\) by an arbitrary product containing all the problem's relevant variables raised to as yet unspecified powers, i.e. \(x^\alpha y^\beta z^\gamma \sim 1\). We call this product our scaling relation, because it shows how each variable will change as the scale (magnitude) of the other variables are changed.

Now, our scaling relation is composed of different factors, each of which has possibly different combinations of the basic dimensions (length, time, mass, charge, et cetera). Suppose we have a system whose relevant variables are known to be position \(r\), velocity \(v\), and time \(t\), so that our scaling relation is given by \(r^\alpha v^\beta t^\gamma \sim 1\). We denote the dimensions of the variable \(x\) by \(\left[x\right]\), and we have \(\left[r\right] = \text{L}\), \(\left[v\right] = \text{L}\text{T}^{-1}\), \(\left[t\right] = \text{T}\). If we replace our variables by their basic dimensions, we have \(\text{L}^\alpha \text{L}^\beta\text{T}^{-\beta} \text{T}^\gamma \sim 1\), or \(\text{L}^{\alpha+\beta} \text{T}^{-\beta + \gamma} \sim 1 = \text{L}^0\text{T}^0\).

Thus, we find that the two equations \(\alpha + \beta = 0\) and \(-\beta + \gamma = 0\) govern the behavior of our scaling relation so that \(-\alpha = \beta = \gamma\). As we'll see, we can pick an arbitrary value for one of the undetermined exponents and obtain the value of the other two. Choosing \(\alpha = -1\), we have \(\beta=\gamma=1\) so that our dimensionless relation becomes \(r^{-1}vt \sim 1\), which leads to \(r \sim vt\), the usual relation between position and velocity. Thus, without knowing anything about kinematics, we were able to obtain a defining relation solely using dimensional analysis. Had we picked another value for \(\alpha\), say \(\frac65\), we'd still end up with the same scaling relation since we'd take the \(\sqrt[\frac65]{r}\) to obtain an integer power for a variable of interest. The value we choose is inconsequential, the scaling conditions demand certain relative values for the exponents, and it is their relative values that govern the system.

The technique might not yet seem all that impressive, and \(r=vt\) is perhaps the simplest relation in all of physics and is known intuitively to anyone who's ever driven a car, or waited for a train. However, the dimensional analysis technique is useful far beyond such a simplistic example and can produce results in situations where intuition falls flat. All that's required of the user is to determine which physical quantities should be important, and unimportant, for the phenomenon under consideration.

Atomic bomb blast radius:

strangesounds

One famous triumph of dimensional analysis was in finding a formula for the blast radius of an exploding atomic bomb as a function of time. Using this scaling relation, it was possible to make accurate estimates of the (highly classified) total energy of the US atomic weapon arsenal by looking at photographs of the blast released to newspapers.

Let's find the relation for ourselves. Our first task is to identify the relevant physical quantities. For one, we are interested in the total energy of the bomb \(E\), which has units of \(\left[E\right] = \text{M}\text{L}^2\text{T}^{-2}\) that must be one of our quantities. We're also interested in the radius of the blast \(r\), and the time since the explosion \(t\), which have units of \(\text{L}\) and \(\text{T},\) respectively. So they too must be included in our scaling relation.

Thus far, we have

explosive energy of the bomb \(E\)

blast front radius \(r\)

time since explosion \(t.\)

It seems we have yet to complete our set of scaling variables. For one, we should certainly expect a more rapid expansion of the shock wave when the bomb explodes in air, compared to when it explodes in an ocean of molasses. For another, we notice that our scaling variables currently include only one variable with a unit of mass. We need all dimensions to drop out in our relation, so we cannot have the bomb energy \(E\) without another mass-dependent quantity.

What we need to complete our set is the density of the medium into which the bomb explodes, \(\rho\), which has units of \(\text{M}\text{L}^{-3}\).

Solving the exponent relations, we find \(\alpha = - \beta\), \(\delta = 2\beta\), and \(5\beta = -\gamma\). We thus have three equations in four exponent variables, so we are free to choose the value of one of them. Choosing \(\alpha = 1\), we find \(\beta = -1\), \(\delta = 2\), and \(\gamma = 5\), which makes our scaling relation

\[\rho E^{-1} r^5 t^2 \sim 1.\]

Thus the radius of the bomb front as a function of time must scale as

\[r \sim \sqrt[5]{\frac{Et^2}{\rho}},\]

i.e.

\[r \propto t^{2/5}.\]

This is quite remarkable. Without knowing a single thing about the physics of bomb explosions under heavy atmospheres, we are able to obtain the correct scaling behavior for a bomb front as it explodes into a massive medium.

Challenge yourself with the following problem:

Using the following time-stamped photographs, estimate the explosive energy of the atomic bomb used in the Trinity tests.

Buckingham \(\pi\) Theorem

Now that we've used dimensional analysis to solve a real world problem, it's a good time to formalize our approach and say a bit more about what's going on. So far, our procedure has been to identify \(n\) physical variables \(a,b,c,\ldots\), form a scaling product \(a^\alpha \times b^\beta \times c^\delta \times \cdots\), and substitute the \(m\) independent dimensions of our variables to form a system of \(m\) equations in \(n\) variables. Solving these equations yields the exponent values needed to form a dimensionless scaling product, \(\pi \sim 1\).

For example, in the atomic bomb problem, we have the following system:

We can write this equation as \(\mathbf{D}\vec{e}=0\), where \(\mathbf{D}\) denotes the \(n \times n\) matrix of our variables' dimensions, and \(\vec{e}\) denotes the length-\(m\) vector of our variables' exponents in the scaling relation.

As yet, this procedure has yielded unique solutions that lead to a single dimensionless product. In general, however, such a system will yield a total of \(n-m\) independent solutions, so we can have multiple dimensionless scaling products \(\{\pi_1, \pi_2\}\). In such cases, the analysis no longer implies \(\pi_i \sim 1\), but instead \(g(\pi_1)=f(\pi_2),\) where \(f\) and \(g\) are arbitrary functions. For simplicity, we can say that we have \(\pi_1 = f(\pi_2)\), and in general \(\pi_1 = f(\pi_2, \ldots, \pi_n)\).

One of the great transformations in physics was the move from Newtonian gravity to the more powerful theory of general relativity. For a time, there was controversy as to whether or not the new theory was correct. One of the major steps in confirming the new theory was in the prediction of the deflection of starlight by the gravitational field of the Sun.

If one assigns the photon a mass of \(\frac Ec = \frac{hf}c\), then it can be shown that in passing the Sun, the photon is deflected through the angle \(\theta_\text{Newton}\). In contrast, general relativity predicts that the bending is instead exactly twice the Newtonian deflection, \(\theta_\text{GR} = 2\theta_{\text{Newton}}\). We won't be able to resolve the controversy, but we should be able to estimate the dependence of \(\theta_\text{Newton}\) on relevant parameters like the speed of light, the mass of the Sun, and the gravitational constant \(G\).

First of all, we'd expect the bending of light to change magnitude if the gravitational constant \(G\), the density of the Sun \(\rho_\text{Sun}\), or the radius of the Sun \(R_\text{Sun}\) were to change. If a classical particle were to pass through the Sun's gravitational field, we'd expect its energy, through its velocity, to have an impact on the kind of orbit it would have in the vicinity of the Sun (whether it enters an orbit, gets shifted, or spirals into the Sun's surface), and therefore it is reasonable to expect that the velocity of light \(c\) should come into play.

We don't expect radiation or the light's frequency to play an important role, and therefore we can probably leave out Boltzmann's constant \(k_B\), as well as Planck's constant \(h\).

The units of \(\rho\), as before, are \(\text{M}\text{L}^{-3}\). The speed of light \(c\) gets us units of \(\text{L}\text{T}\), and the radius of the Sun \(R\) gives \(\text{L}\). Now, the gravitational constant has units such that

\[G\frac{m^2}{r^2}\]

yields units of force, or \(\text{M}\text{L}\text{T}^{-2}\), and thus we have \(\left[G\right] = \text{M}^{-1}\text{L}^3\text{T}^{-2}\).

Finally, we have our variable of interest \(\theta,\) which is already a dimensionless parameter. As we have five variables \((n=5)\) and three basic dimensions \((m=3),\) we expect \(n - m = 2\) dimensionless products such that \(\pi_1 = \theta = f(\pi_2) = f\big(G^\alpha \rho^\beta R^\gamma c^\delta\big)\).

The simplicity of dimensional analysis is also its curse. As we now need to relate two dimensionless quantities, \(\pi_2\) can be the argument for almost any function we can imagine, i.e. \(\sin\), \(\cos\), \(\tanh^{-1}\), \(\exp\), et cetera. However, we can use our intuition to narrow down the possibilities. First, if \(G\), \(\rho\), or \(R\) increases, we'd expect the deflection to increase, and we have the inverse expectation on the speed of light \(c\) (\(c\) is obviously a constant, but should it increase, we'd expect the impact of the Sun on photon trajectory to be diminished, in theory).

Now, if we Taylor-expand the function \(f\) in \(\pi_2\), we'd expect no constant term, i.e. if \(G\), \(\rho\), or \(R\) is zero, or \(c\) is infinite, then we'd expect no deviation whatsoever, and therefore \(f(0) = 0\). Now, the second term in the expansion should be the derivative of \(f\) at zero multiplied by \(\pi_2\). Assuming \(f^\prime(0)\neq 0\), we would have \(\theta \sim \pi_2 + O\left(\pi_2^2\right) \sim \frac{G\rho R^2}{c^2}\). Now, in fact, this is the correct scaling relation between \(\theta\) and the physical quantities \(G, \rho, R\) and \(c\).

hubble-lensing

We don't pick up any insight as to whether \(\theta \approx \theta_\text{Newton}, 4\theta_\text{Newton}\) or \(10^6\theta_\text{Newton}\), but we can say that the light bending depends on the physical parameters according to the scaling relation we've obtained. Above, we see a dramatic example of light bending through a gravitational field, forming an Einstein ring.

Note: We had the choice to bring the mass of the gravitating object into the problem either through the density of the object \(\rho\) or by explicitly including the mass of the object as one of our physical quantities. We chose to use the density. The reason for this is that the density is a scale-free quantity, whereas the mass varies with the radius of the object. Were we to have used \(M\), we would needlessly complicate our analysis, having two quantities that each scale with \(R\). Using \(\rho\) makes the dependence on planet size transparent. In general, whenever there are multiple ways to represent a given quantity, one of which is scale-free, it is usually best to pick the scale-free form.

A large, spherical, and very diffuse cloud of molecular hydrogen sits in empty space. The radius of the cloud is \(r_0\) and the molecules are initially at rest. Let \(t_0\) be the time it takes for this cloud to completely collapse, i.e. the time it takes for a molecule at \(r_0\) to free-fall into the origin. Similarly, let \(t_1\) be the time it takes for a cloud of radius \(r_1=4r_0\) to collapse. What is \(\frac{t_1}{t_0}?\)

\(\) Details and Assumptions:

The entire cloud is collapsing in this process.

Both clouds have the same mass.

Escape Velocity

Use dimensional analysis to show that the velocity required for a massive object to escape the gravitational field of a planet scales as \(v \sim \sqrt{\frac{GM}R}\).

Shortcomings of Dimensional Analysis

The benefits of dimensional analysis are clear. The technique is extremely simple to use and can give us physical intuition where none exists, to begin with. With its simplicity comes a number of shortcomings. Although we can usually, very quickly, get to the scaling solution for a problem, dimensional analysis can never yield exact answers in cases of numerical prefactors. In cases where we end up with multiple dimensionless products \(\{\pi_1, \pi_2,\ldots\}\) such that \(\pi_1 \sim f(\pi_2, \ldots)\), we need to come up with physical reasons to narrow down the form of \(f\), and take limiting cases to demonstrate the plausibility of a proposed solution.

However, as one of the use cases for dimensional analysis is a situation where we don't know much about a problem, this can be a potentially tall order. Moreover, the method works best when the answer we're after is a simple product. When the answer involves the sum of variables of the same dimension, things can get hairy. In these cases, we may be able to find several limiting cases (when one or the other summand goes to zero) and then stitch the solutions together afterward.

However, both as a learning and a research tool, dimensional analysis is a powerful weapon and can be the first and last line of attack on new problems.

Once water crosses from xylem to phloem, it flows through the trunk toward the energy consuming parts of the tree. The water becomes solvent for sucrose, and the flowing material in phloem is sap. The phloem is essentially one long pipe of constant radius, with length equal to the height of the tree.

Compared to water, sap is significantly more viscous and thus will exert friction on itself that slows its flow. As with the membrane, we can find a relation between osmotic pressure and volume flow that we call the hydraulic resistance of the sap in the stem, \(\mathcal{R}_{\text{stem}}\).

We can calculate this resistance exactly but it is a little involved, so we'll first try to derive its functional form by scaling arguments. As the great John Wheeler once said, "Never make a calculation until you know the answer."

We expect the area of the cross section, the length of the pipe \(h\), the pressure across the pipe \(\Delta p\), and the viscosity of the liquid \(\eta\) to come into play.